Evaluate: $\int$ sin x sin 2x sin 3x dx

It is known that,  sin A sin B = $\frac{1}{2}$ cos A - B - cos A + B

$\therefore \int \sin \mathrm{x} \sin 2 \mathrm{x} \sin 3 \mathrm{x} \mathrm{dx} = \int \left[ \sin \mathrm{x} \times \frac{1}{2} \cos 2 \mathrm{x} - 3 \mathrm{x} - \cos 2 \mathrm{x} + 3 \mathrm{x} \right]$

                                        $$\begin{gathered} = \frac{1}{2} \int \sin \mathrm{x} \cos ( - \mathrm{x} ) - \sin \mathrm{x} \cos 5 \mathrm{x} \mathrm{dx} \\ = \frac{1}{2} \int \sin \mathrm{x} \cos \mathrm{x} - \sin \mathrm{x} \cos 5 \mathrm{x} \mathrm{dx} \\ = \frac{1}{2} \int \frac{\sin 2 \mathrm{x}}{2} \mathrm{dx} - \frac{1}{2} \int \sin \mathrm{x} \cos 5 \mathrm{x} \mathrm{dx} \\ = \frac{1}{4} \left[ \frac{ - \cos 2 \mathrm{x} }{2} \right] - \frac{1}{2} \int \frac{1}{2} \sin \mathrm{x} + 5 \mathrm{x} + \sin \mathrm{x} - 5 \mathrm{x} \mathrm{dx} \\ = \frac{ - \cos 2 \mathrm{x}}{8} - \frac{1}{4} \int \left( \sin 6 \mathrm{x} + \sin ( - 4 \mathrm{x} ) \right) \mathrm{dx} \end{gathered}$$

                                        $$\begin{gathered} = \frac{ - \cos 2 \mathrm{x}}{8} - \frac{1}{4} \left[ \frac{- \cos 6 \mathrm{x}}{6} + \frac{\cos 4 \mathrm{x}}{4} \right] + \mathrm{C} \\ = \frac{ - \cos 2 \mathrm{x}}{8} - \frac{1}{8} \left[ \frac{- \cos 6 \mathrm{x}}{3} + \frac{\cos 4 \mathrm{x}}{2} \right] + \mathrm{C} \\ = \frac{ - \cos 2 \mathrm{x}}{8} - \frac{1}{8} \left[ \frac{2 \times (- \cos 6 \mathrm{x} ) + 3 \times \cos 4 \mathrm{x} }{3 \times 2} \right] + \mathrm{C} \\ = \frac{ - \cos 2 \mathrm{x}}{8} - \frac{1}{8} \left[ \frac{- 2 \cos 6 \mathrm{x} + 3 \cos 4 \mathrm{x} }{6} \right] + \mathrm{C} \\ = \frac{6 \times ( - \cos 2 \mathrm{x} )}{6 \times 8} - \frac{1}{8 \times 6} \left[ - 2 \cos 6 \mathrm{x} + 3 \cos 4 \mathrm{x} \right] + \mathrm{C} \\ = \frac{ - 6 \cos 2 \mathrm{x}}{48} - \frac{1}{48} \left[ - 2 \cos 6 \mathrm{x} ) + 3 \cos 4 \mathrm{x} \right] + \mathrm{C} \end{gathered}$$

                                        $$\begin{gathered} = \frac{1}{48} \left[ - 6 \cos 2 \mathrm{x} - ( - 2 \cos 6 \mathrm{x} + 3 \cos 4 \mathrm{x} ) \right] + \mathrm{C} \\ = \frac{1}{48} \left[ 2 \cos 6 \mathrm{x} - 3 \cos 4 \mathrm{x} - 6 \cos 2 \mathrm{x} ) \right] + \mathrm{C} \end{gathered}$$